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Mendel crossed peas having round seeds and yellow cotyledons (seed leaves) with peas having wrinkled seeds and green cotyledons. All the \(F_{1}\) plants had round seeds with yellow cotyledons. Diagram this cross through the \(\mathrm{F}_{2}\) generation, using both the Punnett square and forked-line, or branch diagram, methods.

Short Answer

Expert verified
In Mendel's pea experiment, given the traits for seed shape (round or wrinkled) and cotyledon color (yellow or green), we determined the F2 generation genotypes using both Punnett square and forked-line methods. The F2 generation genotypes are distributed as follows: 1 RRYY (round seeds, yellow cotyledons), 2 RrYY (round seeds, yellow cotyledons), 1 rrYY (wrinkled seeds, yellow cotyledons), 2 RRYy (round seeds, green cotyledons), 4 RrYy (round seeds, green cotyledons), 2 rrYy (wrinkled seeds, green cotyledons), 1 RRyy (round seeds, green cotyledons), 2 Rryy (round seeds, green cotyledons), and 1 rryy (wrinkled seeds, green cotyledons).

Step by step solution

01

Determine the F1 generation genotypes

Using the Punnett square method, we will find the possible genotype combinations for the F1 generation. Since both traits are inherited independently, we only need to cross the genotypes for each trait separately and then combine the results. Seed shape (R and r) cross: ``` R R r Rr Rr r Rr Rr ``` Cotyledon color (Y and y) cross: ``` Y Y y Yy Yy y Yy Yy ``` Combine the two results: all F1 generation plants have the genotype RrYy (round seeds, yellow cotyledons).
02

Set up the F2 generation genotypes

For the F2 generation, we will need to cross the F1 generation plants (RrYy) with themselves. Again, we will first cross the genotypes for seed shape and cotyledon color separately and then combine the results.
03

Determine the F2 generation seed shape genotypes using the Punnett square method

Cross the Rr genotype for seed shape (for F1 plants) with itself: ``` R r R RR Rr r Rr rr ``` We have the genotypes: RR (1), Rr (2), and rr (1) for the F2 generation seed shape.
04

Determine the F2 generation cotyledon color genotypes using the Punnett square method

Cross the Yy genotype for cotyledon color (for F1 plants) with itself: ``` Y y Y YY Yy y Yy yy ``` We have the genotypes: YY (1), Yy (2), and yy (1) for the F2 generation cotyledon color.
05

Combine the F2 generation results using the forked-line method

Now we can combine the results from steps 3 and 4 using the forked-line method. * Seed shape: 1 (RR) : 2 (Rr) : 1 (rr) * Cotyledon color: 1 (YY) : 2 (Yy) : 1 (yy) Using forked-line method: ``` 1 (RRYY) : 2 (RrYY) : 1 (rrYY) // seed shape combined with YY 2 (RRYy) : 4 (RrYy) : 2 (rrYy) // seed shape combined with Yy 1 (RRyy) : 2 (Rryy) : 1 (rryy) // seed shape combined with yy ``` The F2 generation genotypes are distributed as follows: 1 RRYY, 2 RrYY, 1 rrYY, 2 RRYy, 4 RrYy, 2 rrYy, 1 RRyy, 2 Rryy, and 1 rryy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Punnett Square
The Punnett Square is a grid created by English scientist Reginald Punnett to predict the possible genotypes and phenotypes resulting from a genetic cross. This handy tool helps visualize how alleles from each parent can combine during fertilization. It is especially useful for understanding Mendelian genetics, where traits are inherited according to specific dominant and recessive patterns.

In a typical Punnett Square for a single trait, each square represents a potential offspring. You fill in the grid by combining the alleles from either side. For example, when crossing round seeds (R) and wrinkled seeds (r):
  • You write the R alleles across the top.
  • You write the r alleles along the side.
  • Each inner square shows a possible gene combination that the offspring might inherit (e.g., Rr).
For dihybrid crosses, the process can become more complex because you are dealing with two traits simultaneously, requiring a larger square or separate squares for each trait.

The Punnett Square helps us see the probabilities of offspring traits, often revealing dominant traits more prominently than recessive, just like in the case of peas having round seeds and yellow cotyledons—where roundness (R) and yellowness (Y) are both dominant traits.
Forked-line Method
The Forked-line Method, also known as a branch diagram, is an alternative to the Punnett Square for determining the genotype probabilities for offspring, particularly in dihybrid or more complex genetic crosses.

This method simplifies the process by dealing with each trait separately, and then combining the results. Here's how it works:
  • Step 1: Line out the possible outcomes for one trait (e.g., seed shape).
  • Step 2: From each outcome, branch out the possible results for the second trait (e.g., cotyledon color).
  • Step 3: Multiply the probabilities along each branch to get the overall likelihood of each genotype combination.
This branching strategy lets you manage the complexity of multiple traits more efficiently than the Punnett Square. In our pea example, you could separately draw outcomes for seed shape (round, wrinkled) and cotyledon color (yellow, green), then match these, calculating the final probabilities of combinations like RrYy or RRyy through multiplication.
Dihybrid Cross
A dihybrid cross involves organisms that are heterozygous for two different traits. It is based on Gregor Mendel's principle of independent assortment. This principle states that alleles for separate traits are distributed to sex cells (& thereby offspring) independently of one another.

When you conduct a dihybrid cross, you are essentially looking at how two pairs of alleles segregate during genetic recombination. In the classic Mendelian experiment with pea plants, we examine seed shape (R - round, r - wrinkled) and cotyledon color (Y - yellow, y - green).

To perform a dihybrid cross, follow these basic steps:
  • Write down the genotypes of the parents, which often starts as heterozygous in each trait (e.g., RrYy).
  • Apply either the Punnett Square or the Forked-line Method to predict the probabilities of offspring genotypes. With the Punnett Square, you'd typically form a 4x4 grid for two traits.
  • Determine phenotypic ratios. In Mendelian genetics, a typical dihybrid cross will result in a 9:3:3:1 ratio in the phenotype of the offspring when both parents are heterozygous for both traits.
This process helps illustrate how genetic variation is maintained across generations, showcasing how traits are combined in offspring ranging from dominant-to-dominant to recessive-to-recessive presentations.

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Most popular questions from this chapter

The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the \(p\) value used to reject or not reject the hypothesis. In the case of the chi- square analysis of genetic crosses, would the use of a standard of \(p=0.10\) be more or less stringent about not rejecting the null hypothesis? Explain.

Albinism in humans is inherited as a simple recessive trait. For the following families, determine the genotypes of the parents and offspring. (When two alternative genotypes are possible, list both.) (a) Two normal parents have five children, four normal and one albino. (b) A normal male and an albino female have six children, all normal. (c) A normal male and an albino female have six children, three normal and three albino. (d) Construct a pedigree of the families in (b) and (c). Assume that one of the normal children in (b) and one of the albino children in (c) become the parents of eight children. Add these children to the pedigree, predicting their phenotypes (normal or albino).

Dentinogenesis imperfecta is a rare, autosomal, dominantly inherited disease of the teeth that occurs in about one in 8000 people (Witkop 1957 ). The teeth are somewhat brown in color, and the crowns wear down rapidly. Assume that a male with dentinogenesis imperfecta and no family history of the disease marries a woman with normal teeth. What is the probability that (a) their first child will have dentinogenesis imperfecta? (b) their first two children will have dentinogenesis imperfecta? (c) their first child will be a girl with dentinogenesis imperfecta?

Albinism, lack of pigmentation in humans, results from an autosomal recessive gene (a). Two parents with normal pigmentation have an albino child. (a) What is the probability that their next child will be albino? (b) What is the probability that their next child will be an albino girl? (c) What is the probability that their next three children will be albino?

A certain type of congenital deafness in humans is caused by a rare autosomal (not X-linked) dominant gene. (a) In a mating involving a deaf man and a deaf woman (both heterozygous), would you expect all the children to be deaf? Explain your answer. (b) In a mating involving a deaf man and a deaf woman (both heterozygous), could all the children have normal hearing? Explain your answer. (c) Another form of deafness is caused by a rare autosomal recessive gene. In a mating involving a deaf man and a deaf woman, could some of the children have normal hearing? Explain your answer.

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